A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equation y(x,t)=0.01msin[(62.8m−1)x]cos[(628s−1)t].Assuming π=3.14, the correct statement(s) is/are
A
The number of nodes is 5.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
the length of the string is 0.25m.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The maximum displacement of the midpoint of the string from its equilibrium position is 0.01m.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The fundamental frequency is 100Hz.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct options are B the length of the string is 0.25m. C The maximum displacement of the midpoint of the string from its equilibrium position is 0.01m. 20π=2πλ λ=110m=0.1m length of the spring =0.5×12=0.25m Mid point is the antinode Frequency at this mode is f=200π2π=100Hz Fundamental frequency =1005=20Hz